Question

Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{x\left( 2 \log x + 1 \right)}{\sin y + y \cos y}\] given that

\[y = \frac{\pi}{2}\] when x = 1.

Answer 1

The given differential equation is \[\frac{dy}{dx} = \frac{x\left( 2\log x + 1 \right)}{\text { sin }y + y\text { cos }y}\]

Separating the variables in equation (1), we get: \[\left( \sin y + y\cos y \right)dy = x\left( 2\log x + 1 \right)dx\]  ...(2)

Integrating both sides of equation (2), we have:

\[\int\left( \sin y + y\cos y \right)dy = \int x\left( 2\log x + 1 \right)dx\] ...(3)

Now, 

\[\int\sin y dy = - \cos y + C\]

\[\in ty\cos y dy = y\sin y + \cos y + C\]  (Using by parts)

∴ \[\int\left( \sin y + y\cos y \right)dy = - \cos y + y\sin y + \cos y + C_1 = y\sin y + C_1\]  ...(4)

\[\text { Let } I = \int\left( 2x\log x + x \right)dx\]                      (using by parts)

\[ = \int2 x\log x dx + \int x dx\]

\[ = 2\left[ \log x\left( \int x dx \right) - \int\left( \frac{d}{dx}\left( \log x \right) . \int x dx \right) dx \right] + \frac{x^2}{2} + C_2 \]

\[ = 2\left[ \log x \times \frac{x^2}{2} - \in t\frac{1}{x} \times \frac{x^2}{2}dx \right] + \frac{x^2}{2} + C_2 \]

\[ = 2\left[ \frac{x^2}{2}\log x - \frac{x^2}{4} \right] + \frac{x^2}{2} + C_2 \]

\[ = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C_2 \]

\[ = x^2 \log x + C_2 . . . \left( 5 \right) \]

Putting the values in equation (3), we get:

\[y\sin y = x^2 \log x + {C, \text { where } C=C}_2 {-C}_1\]          ...(6)

On putting y = \[\frac{\pi}{2}\] and x = 1 in equation (6), we get:

C = \[\frac{\pi}{2}\]

∴ The particular solution of the given differential equation is

\[y\sin y = x^2 \log x + \frac{\pi}{2}\] .